Problem Solving

In a sequence of 40 numbers, each term, except for the first one, is 7 less than the previous term. If the greatest term in the sequence is 281, what is the smallest term in the sequence?

A. 8
B. 7
C. 1
D. 0
E. −6

[spoiler]OA=A[/spoiler].

What is the best approach I can use here? Could anyone give me some help? Please. I'd be thankful.

Hi Gmat_misson.

Let's take a look at your question.

In a sequence of 40 numbers, each term, except for the first one, is 7 less than the previous term. If the greatest term in the sequence is 281, what is the smallest term in the sequence?

We have that $$a_1=281,\ a_2=281-7,\ \ a_3=\left(281-7\right)-7,\ a_4=\left(\left(281-7\right)-7\right)-7$$ $$a_1=281,\ a_2=a_1+\left(-7\right),\ \ a_3=a_1+\left(-14\right),\ a_4=a_1+\left(-21\right),\ ....$$ $$a_1=281,\ a_2=a_1+\left(-7\right),\ \ a_3=a_1+2\cdot\left(-7\right),\ a_4=a_1+\left(3\right)\left(-7\right),\ ....$$ Therefore, we get the general form: $$a_n=a_1+\left(n-1\right)\left(-7\right).$$ So, the last term (nº40, the smallest) is equal to $$a_{40}=a_1+\left(40-1\right)\left(-7\right)=281+\left(39\right)\left(-7\right)=281-273=8.$$ So, the correct answer is the option A.

I hope this explanation is clear enough.

Regards.

Gmat_mission wrote:In a sequence of 40 numbers, each term, except for the first one, is 7 less than the previous term. If the greatest term in the sequence is 281, what is the smallest term in the sequence?

A. 8
B. 7
C. 1
D. 0
E. −6

Since each term is 7 less than the preceding term, the numbers are EVENLY SPACED.
For any evenly spaced set, where d is the distance between successive terms:
$$a_{n} = a_{1} - (n-1)d$$
In the problem above:
aâ‚„â‚€ = ?
a� = 281
n = 40
Since each term is 7 less than the preceding term, d =7.
Plugging these values into the equation above, we get:
aâ‚„â‚€ = 281 - (40-1)(7) = 281 - 273 = 8.

The correct answer is A.